$f(x, y) = (\cos(x + y), -\sin(x - y))$ What is the curl of $f$ at $\left( \dfrac{3\pi}{2}, \dfrac{\pi}{2} \right)$ ?
Answer: The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ -\sin(x - y) \right] \\ \\ &= -\cos(x - y) \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ \cos(x + y) \right] \\ \\ &= -\sin(x + y) \end{aligned}$ Therefore: $\text{curl}(f) = -\cos(x - y) + \sin(x + y)$ The curl of $f$ at $\left( \dfrac{3\pi}{2}, \dfrac{\pi}{2} \right)$ is $1$.